If the length of each of two equal sides of an isosceles triangle is 10 cm and the angle made by these side on the third side is 45°, then find perimeter of the triangle.
Answer & explanation
Correct answer: option 1
The triangle ABC is an isosceles triangle,
\(\angle\)BAC = \(\angle\)BCA = 45°
Therefore, \(\angle\)B = 90°
In ΔABC:
⇒ AC2 = AB2 + BC2
⇒ AC2 = 100 + 100 = 200 cm
⇒ AC = 10 \(\sqrt {2}\) cm
Now,
Perimeter = 10 + 10 + 10\(\sqrt {2}\)
= 10 (2 + \(\sqrt {2}\)cm)
= 10(2 + \(\sqrt {2}\)) cm